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Question: If \(a\cos \theta -b\sin \theta =c\), show that \(a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{...

If acos⁡θ−bsin⁡θ=ca\cos \theta -b\sin \theta =c, show that asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}.

Explanation

Solution

Hint: Square both sides of the given expression and use the trigonometric identities cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta and sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta to obtain the square of the expression on LHS of the second expression. On rearrangement, the only expression that remains on the LHS is the square of asin⁡θ+bcos⁡θa\sin \theta +b\cos \theta and the RHS is an expression in a, b and c. Then find square roots on both sides to get the required result.

Complete step-by-step answer:
We are given the equation acos⁡θ−bsin⁡θ=ca\cos \theta -b\sin \theta =c. The first step in this case will be to square both sides. Thus, squaring both sides gives us

(acos⁡θ−bsin⁡θ)2=c2 ⇒a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ=c2 \begin{aligned} &{{\left( a\cos \theta -b\sin \theta \right)}^{2}}={{c}^{2}} \\\ & \Rightarrow {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\\ \end{aligned}

Now we know that cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta and that sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta . We substitute these values in the squared expression to obtain
a2(1−sin⁡2θ)+b2(1−cos⁡2θ)−2absin⁡θcos⁡θ=c2 ⇒a2+b2−a2sin⁡2θ−b2cos⁡2θ−2absin⁡θcos⁡θ=c2 ⇒a2sin⁡2θ+b2cos⁡2θ+2absin⁡θcos⁡θ=a2+b2−c2                          …(1) \begin{aligned} &{{a}^{2}}\left(1-{{\sin }^{2}}\theta \right)+{{b}^{2}}\left( 1-{{\cos }^{2}}\theta \right)-2ab\sin \theta \cos \theta ={{c}^{2}} \\\ & \Rightarrow {{a}^{2}}+{{b}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta -{{b}^{2}}{{\cos }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\\ & \Rightarrow {{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta ={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \ldots \left( 1 \right) \\\ \end{aligned}

Now, to consider the expression whose value has to be found, we square that expression to obtain
(asin⁡θ+bcos⁡θ)2=a2sin⁡2θ+b2cos⁡2θ+2absin⁡θcos⁡θ{{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta
The expression on this equation’s RHS is the same as the expression on the LHS of equation (1). Hence, the value of (asin⁡θ+bcos⁡θ)2=a2+b2−c2{{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}+{{b}^{2}}-{{c}^{2}}.

Taking square roots on both sides of this equation, we obtain asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}, which is the required result.
Hence the result is proved.

Note: The fairly simple question seems difficult at first sight but the trick is to transform the expression in the question to the one whose value has to be found. To do this in questions involving sin⁡\sin and cos⁡\cos functions of the same angle θ\theta , the best way is to find the square on both sides and then use the identities of cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta and sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta = 1-{{\cos}^{2}}\theta.